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12345 [234]
3 years ago
11

What is the 1/2 half life of Cesium 137 in years?

Chemistry
2 answers:
blondinia [14]3 years ago
8 0

Answer:

About 30.17 years

Explanation:

About 94.6% decays by beta emission to a metastable nuclear isomer of barium: barium-137m . The remainder directly populates the ground state of barium-137, which is stable.

svetoff [14.1K]3 years ago
4 0

Answer:

about 30.17 years

Explanation:

Caesium-137 has a half-life of about 30.17 years. About 94.6% decays by beta emission to a metastable nuclear isomer of barium: barium-137m (137mBa, Ba-137m). The remainder directly populates the ground state of barium-137, which is stable.

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Murrr4er [49]

Answer: think a

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How many atoms are in the 1.5 moles of tin atoms?
Anastasy [175]

Answer:

option C = 9.0 x 10²³ atoms

Explanation:

Data Given:

no. of moles of tin (Sn) atoms = 1.5 moles

no. of tin (Sn) atoms = ?

Solution:

Formula used to find number of atoms

                  no. of moles = no. of atoms / Avogadro's number

Rearrange the above equation:

          no. of atoms =   no. of moles x Avogadro's number . . . . . . (1)

Where

Avogadro's number = 6.022 x 10²³

Put values in equation 1

           no. of atoms = 1.5 x 6.022 x 10²³

           no. of atoms = 9.033 x 10²³

Round the figure = 9.0 x 10²³ atoms

So option C is correct

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3 years ago
Cassady? you there ?
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who da hell is cassady?

Explanation:

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3 years ago
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Bacteria are very small *<br><br> True<br> False
elena-14-01-66 [18.8K]

Of course they are small

Explanation:

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3 years ago
g Tibet (altitude above sea level is 29,028 ft) has an atmospheric pressure of 240. mm Hg. Calculate the boiling point of water
Marina CMI [18]

<u>Answer:</u> The boiling point of water in Tibet is 69.9°C

<u>Explanation:</u>

To calculate the boiling point of water in Tibet, we use the Clausius-Clayperon equation, which is:

\ln(\frac{P_2}{P_1})=\frac{\Delta H}{R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

P_1 = initial pressure which is the pressure at normal boiling point = 1 atm = 760 mmHg      (Conversion factor:  1 atm = 760 mmHg)

P_2 = final pressure = 240. mmHg

\Delta H_{vap} = Heat of vaporization = 40.7 kJ/mol = 40700 J/mol     (Conversion factor: 1 kJ = 1000 J)

R = Gas constant = 8.314 J/mol K

T_1 = initial temperature or normal boiling point of water = 100^oC=[100+273]K=373K

T_2 = final temperature = ?

Putting values in above equation, we get:

\ln(\frac{240}{760})=\frac{40700J/mol}{8.314J/mol.K}[\frac{1}{373}-\frac{1}{T_2}]\\\\-1.153=4895.36[\frac{T_2-373}{373T_2}]\\\\T_2=342.9K

Converting the temperature from kelvins to degree Celsius, by using the conversion factor:

T(K)=T(^oC)+273

342.9=T(^oC)+273\\T(^oC)=(342.9-273)=69.9^oC

Hence, the boiling point of water in Tibet is 69.9°C

3 0
3 years ago
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